Submission #606435

#	Time	Username	Problem	Language	Result	Execution time	Memory
606435		wiwiho	Meetings (IOI18_meetings)	C++14	0 / 100	75 ms	24712 KiB

meetings

#include "meetings.h"

#include <bits/stdc++.h>
#include <bits/extc++.h>

#define StarBurstStream ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define iter(a) a.begin(), a.end()
#define riter(a) a.rbegin(), a.rend()
#define lsort(a) sort(iter(a))
#define gsort(a) sort(riter(a))
#define pb(a) push_back(a)
#define eb(a) emplace_back(a)
#define pf(a) push_front(a)
#define ef(a) emplace_front(a)
#define pob pop_back()
#define pof pop_front()
#define mp(a, b) make_pair(a, b)
#define F first
#define S second
#define mt make_tuple
#define gt(t, i) get<i>(t)
#define tomax(a, b) ((a) = max((a), (b)))
#define tomin(a, b) ((a) = min((a), (b)))
#define topos(a) ((a) = (((a) % MOD + MOD) % MOD))
#define uni(a) a.resize(unique(iter(a)) - a.begin())
#define printv(a, b) {bool pvaspace=false; \
for(auto pva : a){ \
    if(pvaspace) b << " "; pvaspace=true;\
    b << pva;\
}\
b << "\n";}

using namespace std;
using namespace __gnu_pbds;

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;

using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pdd = pair<ld, ld>;
using tiii = tuple<int, int, int>;

const ll MOD = 1000000007;
const ll MAX = 2147483647;

template<typename A, typename B>
ostream& operator<<(ostream& o, pair<A, B> p){
    return o << '(' << p.F << ',' << p.S << ')';
}

ll ifloor(ll a, ll b){
    if(b < 0) a *= -1, b *= -1;
    if(a < 0) return (a - b + 1) / b;
    else return a / b;
}

ll iceil(ll a, ll b){
    if(b < 0) a *= -1, b *= -1;
    if(a > 0) return (a + b - 1) / b;
    else return a / b;
}

int n;

struct Line{
    ll a = 0, b = 0;
    int l = 1, r = n;
    ll operator()(ll x){
        return a * x + b;
    }
};

ostream& operator<<(ostream& o, Line ln){
    return o << '(' << ln.a << ',' << ln.b << ',' << ln.l << ',' << ln.r << ')';
}

int intersection(Line l1, Line l2){
    return iceil(l1.b - l2.b, l2.a - l1.a);
}

struct DynamicHull{
    deque<Line> dq;

    void insert(Line ln){
        while(!dq.empty() && dq.back()(dq.back().l) >= ln(dq.back().l)){
            dq.pob;
        }
        if(dq.empty()){
            dq.eb(ln);
            return;
        }
        if(dq.back()(dq.back().r) < ln(dq.back().r)){
            if(dq.back().r < n){
                ln.l = dq.back().r + 1;
                if(ln.l > ln.r) return;
                dq.eb(ln);
            }
        }
        if(dq.back().a == ln.a) return;
        int mid = intersection(dq.back(), ln);
        dq.back().r = mid - 1;
        ln.l = mid;
        if(ln.l > ln.r) return;
        dq.eb(ln);
    }

    ll query(int x){
        while(!dq.empty() && dq.front().r < x) dq.pof;
        if(dq.empty()) return 1LL << 60;
        return dq.front()(x);
    }

};

#define lc 2 * id + 1
#define rc 2 * id + 2

struct SegmentTree{
    vector<DynamicHull> st;
    void init(){
        st.resize(4 * n);
    }
    void modify(int x, Line ln, int L = 1, int R = n, int id = 0){
        st[id].insert(ln);
        if(L == R){
            return;
        }
        int M = (L + R) / 2;
        if(x <= M) modify(x, ln, L, M, lc);
        else modify(x, ln, M + 1, R, rc);
    }

    ll query(int l, int r, int x, int L = 1, int R = n, int id = 0){
        if(l <= L && R <= r){
            return st[id].query(x);
        }
        int M = (L + R) / 2;
        if(r <= M) return min(st[id].query(x), query(l, r, x, L, M, lc));
        else if(l > M) return min(st[id].query(x), query(l, r, x, M + 1, R, rc));
        else{
            return min({st[id].query(x), query(l, r, x, L, M, lc), query(l, r, x, M + 1, R, rc)});
        }
    }
};

vector<ll> minimum_costs(vector<int> H, vector<int> L, vector<int> R){
    n = H.size();
    int q = L.size();

    vector<ll> h(n + 1);
    for(int i = 0; i < n; i++) h[i + 1] = H[i];
    for(int i = 0; i < q; i++) L[i]++, R[i]++;

    vector<ll> sum(n + 1), lsum(n + 2), rsum(n + 2);
    vector<int> lp(n + 1), rp(n + 1);
    vector<vector<int>> qry(n + 1);

    vector<ll> ans(q, MAX);
    vector<int> lb(q, -1), rb(q, -1);

    for(int i = 0; i < q; i++){
        qry[R[i]].eb(i);
    }

    auto init = [&](int v, vector<ll>& ssum, vector<int>& p){
        //cerr << "init " << v << "\n";
        deque<int> dq;
        dq.eb(v == 1 ? 0 : n + 1);
        ll tmp = 0;
        for(int i = v == 1 ? 1 : n; 1 <= i && i <= n; i += v){
            while(dq.size() > 1 && h[dq.back()] < h[i]){
                int t = dq.back();
                dq.pob;
                tmp -= abs(t - dq.back()) * h[t];
            }
            tmp += abs(i - dq.back()) * h[i];
            //cerr << "test " << i << " " << tmp << "\n";
            ssum[i] = tmp;
            p[i] = dq.back();
            dq.eb(i);
        }
    };
    init(1, lsum, lp);
    init(-1, rsum, rp);
    for(int i = 1; i <= n; i++){
        sum[i] = lsum[i] + rsum[i] - h[i];
    }

    /*printv(sum, cerr);
    printv(lsum, cerr);
    printv(rsum, cerr);
    printv(lp, cerr);
    printv(rp, cerr);*/

    auto solve = [&](){
        /*cerr << "solve\n";
        printv(h, cerr);
        printv(sum, cerr);*/
        vector<ll> tans = sum;
        deque<int> dq;
        SegmentTree st;
        st.init();
        for(int i = 1; i <= n; i++){
            while(!dq.empty() && h[dq.back()] <= h[i]){
                tans[i] = min(tans[i], tans[dq.back()]);
                dq.pob;
            }
            dq.eb(i);
            int t = rp[i];
            st.modify(i, Line({h[i], tans[i] - t * h[i] - rsum[t] + h[i], 0, t - 1}));
            /*cerr << "test " << i << " : ";
            printv(dq, cerr);
            cerr << t << " " << tans[i] << " " << rsum[t] << "\n";
            cerr << "add " << h[i] << " " << tans[i] - t * h[i] - rsum[t] + h[i] << " " << 0 << " " << t - 1 << "\n";*/
            
            for(int j : qry[i]){
                int l = L[j];
                rb[j] = *lower_bound(iter(dq), l);
                if(rb[j] == i) continue;
                //cerr << "query " << j << " : " << l << " " << rb[j] << "\n";
                ll tmp = st.query(rb[j] + 1, i, i) - lsum[rb[j]] + (rb[j] - l + 1) * h[rb[j]];
                /*cerr << "ok\n";
                cerr << st.query(rb[j] + 1, i, i) << " " << lsum[rb[j]] << " " << rb[j] - l + 1 << " " << h[rb[j]] << " " << tmp << "\n";*/
                ans[j] = min(ans[j], tmp);
            }
        }
        /*cerr << "ok\n";
        printv(ans, cerr);*/
    };

    //cerr << "left to right\n";
    solve();
    
    reverse(h.begin() + 1, h.begin() + n + 1);
    reverse(sum.begin() + 1, sum.begin() + n + 1);
    reverse(lsum.begin() + 1, lsum.begin() + n + 1);
    reverse(rsum.begin() + 1, rsum.begin() + n + 1);
    lsum.swap(rsum);
    
    rp.swap(lp);
    for(int i = 1; i <= n; i++) rp[i] = n - rp[i] + 1;
    reverse(rp.begin() + 1, rp.begin() + n + 1);
    fill(iter(qry), vector<int>());
    L.swap(R);
    for(int i = 0; i < q; i++){
        L[i] = n - L[i] + 1;
        R[i] = n - R[i] + 1;
        qry[R[i]].eb(i);
    }
    lb.swap(rb);

    /*cerr << "right to left\n";
    printv(h, cerr);
    printv(sum, cerr);
    printv(lsum, cerr);
    printv(rsum, cerr);
    printv(rp, cerr);*/
    solve();

    lsum.swap(rsum);
    reverse(h.begin() + 1, h.begin() + n + 1);
    reverse(sum.begin() + 1, sum.begin() + n + 1);
    reverse(lsum.begin() + 1, lsum.begin() + n + 1);
    reverse(rsum.begin() + 1, rsum.begin() + n + 1);
    for(int i = 0; i < q; i++){
        L[i] = n - L[i] + 1;
        R[i] = n - R[i] + 1;
    }
    L.swap(R);
    lb.swap(rb);
    for(int i = 0; i < q; i++) lb[i] = n - lb[i] + 1;

    fill(iter(qry), vector<int>());
    for(int i = 0; i < q; i++) qry[rb[i]].eb(i);

    {
        vector<int> st;
        for(int i = 1; i <= n; i++){
            while(!st.empty() && sum[st.back()] >= sum[i]) st.pob;
            st.eb(i);
            for(int j : qry[i]){
                int l = lb[j];
                int id = *lower_bound(iter(st), l);
                //cerr << "test " << i << " " << j << " " << l << "\n";
                ll tmp = sum[id] - lsum[l] + h[l] * (l - L[j] + 1) - rsum[i] + h[i] * (R[j] - i + 1);
                ans[j] = min(ans[j], tmp);
            }
        }
    }

    return ans;
}

• Subtask 1: 0 / 4
• Subtask 2: 0 / 15
• Subtask 3: 0 / 17
• Subtask 4: 0 / 24
• Subtask 5: 0 / 40